Find speed of two objects after collision

In summary: And then it would be true only for the x direction. The y direction would have a different equation.P=m*v = 0.226kg * 5.59m/s = 1.263314 kg*m/sCorrect.Vay = Va sin 37 -- for A's Y directionVax = Va cos 65 -- for A's X directionVbx = Vb cos 37 -- for B's X directionVby = Vb sin 65 -- For B's Y DirectionAlso correct. The suffixes a, b, x and y are a smart move. Helpful to use something like 1 for before the collision and 2 for after.
  • #1
JJones_86
72
0

Homework Statement


107mnfc.jpg
Puck A has a mass of 0.226 kg and is moving along the x-axis with a velocity of 5.59 m/s. It makes a collision with puck B, which has a mass of 0.452 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of

a) Puck A

b) Puck B

Homework Equations


No idea



The Attempt at a Solution


No Idea


------------
Where do I even start?
 
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  • #2
You have to show some work in order to get help here. Those are the rules. "No idea" doesn't cut it, sorry.

You have to have SOME thoughts on the problem. What have you tried? What concepts does the problem involve?
 
  • #3
G01 said:
You have to show some work in order to get help here. Those are the rules. "No idea" doesn't cut it, sorry.

You have to have SOME thoughts on the problem. What have you tried? What concepts does the problem involve?

Well I do have some thoughts...
I know that it has to do with momentum, and directions.

I really just don't know where to start.. After I find the momentum, where would I go from there?
 
  • #4
Consider the relationship between the momenta before and after collision ...
 
  • #5
catkin said:
Consider the relationship between the momenta before and after collision ...

Will the both have the same momentum?
 
  • #6
That's what the law of conservation tells us ... and there are more than two (you wrote "both") momenta to consider.
 
  • #7
catkin said:
That's what the law of conservation tells us ... and there are more than two (you wrote "both") momenta to consider.

Ok, so how would you figure out what the momentum of the two are after the collision?
 
  • #8
p = mv

Both p (momentum) and v (velocity) are vectors so it's a vector equation.

Start with the horizontal momenta, one for A before and one each for A and B after. After must be the same as before (conservation!). You don't know the magnitude of the velocities of A and B after so you'll have to call them Va and Vb (or some such) and find another equation to find their values.

I've got to go now.
 
  • #9
catkin said:
p = mv

Both p (momentum) and v (velocity) are vectors so it's a vector equation.

Start with the horizontal momenta, one for A before and one each for A and B after. After must be the same as before (conservation!). You don't know the magnitude of the velocities of A and B after so you'll have to call them Va and Vb (or some such) and find another equation to find their values.

I've got to go now.

Ok, so how do you find out the velocity of A & B?

So far I've come up with these equations:

P=m*v
= 0.226kg * 5.59m/s
= 1.263314 kg*m/s

Vay = Va sin 37 -- for A's Y direction
Vax = Va cos 65 -- for A's X direction
Vbx = Vb cos 37 -- for B's X direction
Vby = Vb sin 65 -- For B's Y Direction

So Va = Sqrt((Vax)^2+(Vay)^2)

Just lost on how to find the velocity of Va and Vb...
 
  • #10
I'm not 100% sure but I think Ma*Va+Mb*Vb=Ma*5.59
 
  • #11
Sakha said:
I'm not 100% sure but I think Ma*Va+Mb*Vb=Ma*5.59

That just tells me the Vb is 0...
 
  • #12
JJones_86 said:
P=m*v
= 0.226kg * 5.59m/s
= 1.263314 kg*m/s
Correct.

JJones_86 said:
Vay = Va sin 37 -- for A's Y direction
Vax = Va cos 65 -- for A's X direction
Vbx = Vb cos 37 -- for B's X direction
Vby = Vb sin 65 -- For B's Y Direction
Also correct. The suffixes a, b, x and y are a smart move. Helpful to use something like 1 for before the collision and 2 for after.

JJones_86 said:
So Va = Sqrt((Vax)^2+(Vay)^2)
Correct but not useful in this problem.

Let M stand for the mass of A, 0.226 kg. How many M is the mass of B?

Slightly modifying what you wrote above, the total momentum in the x direction before the collision can be written
P1x = (V1ax * m1) + (V1bx * mb)
= (5.59 * M) + (0 * 2M)
= 5.59M

Use your expressions for velocities after the collision (V2ax etc.) in an expression for the total momentum in the x direction after the collision:
P2x = ...

What is the total momentum in the y direction before and after the collision?
P1y = ...
P2y = ...
 
  • #13
Sakha said:
I'm not 100% sure but I think Ma*Va+Mb*Vb=Ma*5.59
That's only correct if the + sign indicates vector addition.
 

1. How do you calculate the speed of two objects after a collision?

The speed of two objects after a collision can be calculated using the conservation of momentum equation, which states that the total momentum before the collision is equal to the total momentum after the collision. This equation is represented as:
m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the objects and v1, v2, v1', and v2' are the initial and final velocities of the objects, respectively.

2. What information do I need to determine the speed of two objects after a collision?

To determine the speed of two objects after a collision, you will need to know the masses of the objects, as well as their initial velocities before the collision and their final velocities after the collision. This information can be obtained through experimentation or given in a problem.

3. Can the speed of two objects after a collision be greater than their initial speeds?

Yes, the speed of two objects after a collision can be greater than their initial speeds. This is possible if one or both of the objects experience an external force during the collision, causing their velocities to increase.

4. How does the angle of collision affect the speed of two objects after a collision?

The angle of collision can affect the speed of two objects after a collision by changing the direction of their velocities. The conservation of momentum equation still applies, but it must be broken down into its x and y components to account for the change in direction.

5. Is there a difference in calculating the speed of two objects after an elastic collision versus an inelastic collision?

Yes, there is a difference in calculating the speed of two objects after an elastic collision versus an inelastic collision. In an elastic collision, the kinetic energy of the system is conserved, so the objects will have the same speed after the collision as they did before. In an inelastic collision, some kinetic energy is lost and converted into other forms of energy, resulting in the objects having a lower speed after the collision.

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